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2 years ago

What is the value of the expression below when y = 8 , y^2 - y - 2 ? Please answer !!!!! Will mark brainliest !!!!!!!!!!

Mathematics

2 answers:

Lesechka [4]2 years ago

6 0

8^2-8-2 is the equation (I think)
So it’s 54?

Kazeer [188]2 years ago

5 0

Answer: 54.............

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Suppose that 10% of all homeowners in an earthquake-prone area of California are insured against earthquake damage. Four homeown

guapka [62]

Remainder of question:

Find the probability distribution of x

Answer:

The random variable x is defined as: X = {0, 1, 2, 3, 4}

The probability distribution of X:

P(X = 0) = 0.656

P(X = 1) = 0.2916

P(X= 2) = 0.0486

P(X=3) = 0.0036

P(X = 4) = 0.0001

Step-by-step explanation:

Sample size, n = 4

Random variable, X = {0, 1, 2, 3, 4}

10% (0.1) of the homeowners are insured against earthquake, p = 0.1

Proportion of homeowners who are not insured against earthquake, q = 1 - 0.1

q = 0.9

Probability distribution of x,

$P(X = r) = ^nC_r *p^r q^{n-r} \\\\P(X= 0) =(^4C_0 *p^1 q^4 )\\P(X=0) = (^4C_0 *0.1^0 0.9^4 ) = 0.656\\P(X= 1)= (^4C_1 *p^1 q^3 )\\P(X=1) = (^4C_1 *0.1^1 0.9^3 ) = 0.2916\\P(X= 2)=( ^4C_2 *p^2 q^2) \\P(X=2) = (^4C_2 *0.1^2 0.9^2 ) = 0.0486\\P(X= 3) = (^4C_3 *p^3 q^3) \\ P(X=3) = (^4C_3 *0.1^3 0.9^1 ) = 0.0036\\P(X= 4) = (^4C_4 *p^4 q^0 )\\ P(X=4) =(^4C_4 *0.1^4 0.9^0 ) = 0.0001$

5 0

3 years ago

The cost of a home is finaced with a $120,000 30-year fixed rate mortage at 4.5%

almond37 [142]

Answer:

$162,000

Step-by-step explanation:

4.5 of 120,000 is 5400

multiply 5400 by 30 and you get your answer.

7 0

2 years ago

Read 2 more answers

I guess I'm lacking in differential equations. I couldn't solve this question. Can you help me?

Sonja [21]

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Reciprocals

Algebra II

Log/Ln Property: $ln(\frac{a}{b} ) = ln(a) - ln(b)$

Calculus

Derivatives

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Derivative of Ln: $\frac{d}{dx} [ln(u)] = \frac{u'}{u}$

Step-by-step explanation:

Step 1: Define

$ln(\frac{2x-1}{x-1} )=t$

Step 2: Differentiate

Rewrite: $t = ln(\frac{2x-1}{x-1})$
Rewrite [Ln Properties]: $t = ln(2x-1) - ln(x - 1)$
Differentiate [Ln/Chain Rule/Basic Power Rule]: $\frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1$
Simplify: $\frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}$
Rewrite: $\frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}$
Combine: $\frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}$
Reciprocate: $\frac{dx}{dt} = -(2x-1)(x-1)$
Distribute: $\frac{dx}{dt} = (1-2x)(x-1)$